TikTok·Data Scientist·Technical Phone Screen·Intermediate

Intermediate

Jun 2026

Summary

TikTok Data Scientist SQL round, two-part question covering window-function-style aggregation and conditional counting. Pretty straightforward if you know your GROUP BY tricks, but the tie-breaker detail on the first part tripped me up a bit.

Questions Asked (2)

Given a products table with category, price, and discount columns, return the product with the highest discount in each category. If there's a tie on discount, pick the one with the lowest product_id.

Data ModelingProduct Analytics & Metrics

Author's notes

I went straight for a subquery with MAX(discount) grouped by category, joined back to the original table.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the table schema and the exact tie-breaking rules. Then, use a window function like ROW_NUMBER() partitioned by category and ordered by discount DESC, product_id ASC to rank products. Finally, filter to rows where the rank equals 1 to get the top product per category.

Pro tip: Mention that you would validate the result by checking for ties and ensuring the tie-breaker is correctly applied, and discuss performance considerations for large datasets, such as indexing on (category, discount, product_id).

1. Clarify requirements

Confirm the table schema, the definition of 'highest discount' (e.g., discount percentage or amount), and the tie-breaking rule (lowest product_id).

2. Choose the right SQL technique

Decide between using a window function (ROW_NUMBER, RANK, DENSE_RANK) or a correlated subquery. For this problem, ROW_NUMBER with a specific ORDER BY is ideal.

3. Write the query

Construct the SQL: SELECT ... FROM (SELECT *, ROW_NUMBER() OVER (PARTITION BY category ORDER BY discount DESC, product_id ASC) AS rn FROM products) t WHERE rn = 1;

4. Validate and optimize

Test with edge cases (ties, nulls, single product per category) and discuss indexing or alternative approaches for performance.

Key Points to Mention

Use of window functions (ROW_NUMBER, RANK, DENSE_RANK) and their differences
Partitioning by category and ordering by discount DESC, product_id ASC
Handling ties correctly with the tie-breaker
Performance considerations: indexing, avoiding full table scans
Edge cases: null discounts, empty categories, multiple ties
Alternative approaches: correlated subquery, GROUP BY with join

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.

Using a submissions table that tracks employer IDs and whether each submission is buggy or non-buggy, write a query to count both types per employer.

Product Analytics & Metrics

Author's notes

Easy pivot with conditional aggregation.

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AI HintsAI Generated

Suggested Approach

Start by clarifying the table schema and the exact definition of 'buggy' vs 'non-buggy' (e.g., a boolean flag or status column). Then write a single query using conditional aggregation (SUM(CASE WHEN ...) or COUNT(CASE WHEN ...)) grouped by employer_id to produce one row per employer with both counts. Finally, validate the query against edge cases like NULLs or missing values.

Pro tip: Mention that you'd also check for data quality issues (e.g., NULLs in the buggy flag) and consider whether the output should include employers with zero submissions of one type—this shows you think about real-world data, not just the happy path.

1. Clarify the schema and definitions

Ask or state assumptions about the table columns (e.g., employer_id, is_buggy boolean or status string) and confirm what 'buggy' and 'non-buggy' mean. This ensures the query matches the intended logic.

2. Choose conditional aggregation

Use SUM(CASE WHEN is_buggy = TRUE THEN 1 ELSE 0 END) and similarly for non-buggy, or COUNT(CASE WHEN ... THEN 1 END). This avoids multiple queries and returns both counts in one row per employer.

3. Group by employer

Add GROUP BY employer_id so that counts are computed per employer. Optionally order by employer_id or total submissions for readability.

4. Handle edge cases and NULLs

Decide how to treat NULLs in the buggy flag (e.g., exclude them or treat as non-buggy) and ensure employers with no submissions of one type still appear with a count of 0.

5. Validate and explain

Run the query on a small sample or mentally test with edge cases, and be ready to explain the logic and any assumptions you made.

Key Points to Mention

Conditional aggregation using CASE WHEN inside SUM or COUNT
GROUP BY employer_id to get per-employer counts
Handling of NULLs or missing values in the buggy flag
Ensuring both buggy and non-buggy counts appear even if one is zero
Performance considerations: single scan vs multiple subqueries
Clarifying the definition of 'buggy' (boolean, status, etc.)

AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.